A35.1 kg wagon is towed up a hill inclined at 18.3 ◦ with respect to the horizontal. the tow rope is parallel to the incline and has a tension of 125 n in it. assume that the wagon starts from rest at the bottom of the hill, and neglect friction. the acceleration of gravity is 9.8 m/s 2 . how fast is the wagon going after moving 75.4 m up the hill? answer in units of m/s.

4.933m/s

Explanation:

the wagon has a weight of 35.1kg*9.81m/s2 = 343.98N

of that weight 343.98N*sin(18.3)=108N are parallel to the hill and oposit tothe tension of the rope.

then, the force that is moving the wagon is 125N-108N=17N

F=m*a then 17N=35.1kg*a

a=0.4843 m/s2

we have two equations

then

t=10.188s

v=4.933 m/s

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft